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- Physics Initiative in Education
UBC Department of Physics and Astronomy

Deriving Einstein's Mass - Energy
Equation E = mc 2

In this n - Physics Initiative in Education module Einstein's Mass-Energy
Equation E = mc is derived and some of its implications are described.

The pedagogical approach outlined is an example of how a physicist is
able to derive a result "from first principles." This module is an excellent
introduction to Physics 1 1 students to the methodology used by physicists.

The derivation and activities outlined in this module are consistent with
the Prescribed Learning Outcomes set out by the BC Ministry of
Education.

The 7i - Physics Initiative in Education is a program begun in support of the UN
sponsored 2005 International Year of Physics, and Physics Education in the Province of
British Columbia.

This 7t - Physics Initiative in Education Module may be copied and used freely with
attribution to the source and the University of British Columbia. © 2005

Deriving Einstein's Energy (2) jt - Physics Initiative in Education
Expression E = mc 2

L A Little History

Albert Einstein (1879 - 1955) was perhaps one of the most well known public figure of
the 20 th century. Wit
physicist of that period.

the 20 l century. Without question, he was the most widely recognized theoretical

In a brilliant paper titled "Does the Inertia of a Body Depend upon its Energy -Content!"
published in September 1905 in the German physics journal Annalen der Physik, 17,
1905 ("1st die Tragheit eines Korpers von seinem Energiegehalt abhangigl") Einstein
was able to show that the mass of an object is a measure of its energy content.

In this paper he derived an expression that equates mass and energy. The 1905 Mass -
Energy paper was published by Einstein shortly after his "On the Electrodynamics of
Moving Bodies", in which he set out his Special Theory of Relativity. The relationship
between mass and energy is a direct consequence of his Theory of Special Relativity.

Possibly the most widely known physics equation amongst scientists and non-scientists is
Einstein's Mass-Energy Equation

mc 2 (1)

where E is the energy in joules, m is the mass of the object in kilograms and c is the
speed of light in metres per second (c ~ 3.0 x 10 8 m/s ).

What this equation tells us is that if we could convert the mass of an object into energy it
would provide a considerable amount of energy. For instance, the complete conversion
of 1.0 kilogram of mass into energy would generate energy equivalent to the entire daily
energy consumption in Canada. Two hundred kilograms of mass converted into energy
would generate energy equivalent to the daily energy consumption of all people on earth.

Deriving Einstein's Energy (3) jt - Physics Initiative in Education
Expression E = mc 2

It is in the context of the release of energy from the nucleus of the atom that Einstein's
Mass - Energy Equation is best established from an experimental standpoint. Nucelar
energy is released from the nucleus of the atom by nuclear radiation, fission or fusion.

There is a 1906 Gedanken Experiment first proposed by Einstein, and known as the
Einstein Box, can be used to derive the relationship between energy and mass. A
Gedanken Experiment is a thought experiment, a process of deduction reasoning that was
very popular with Einstein and with some of his contemporaries such as Niels Bohr.

In this module a discussion of light pressure followed by a description of the 1906
Gedanken Experiment will be presented, as well as a derivation that uses relativistic
mass. A subsequent section will talk about some misconceptions surrounding Einstein's
Mass - Energy Equation, and will look at the sun, which is a giant fusion reactor.

2, Light Pressure

In about 1600 it was suggested by Johannes Kepler that light pressure from the sun was
why the tail of comets appear to be blown away from the sun. If light quanta carry
momentum and energy then they must exert a light pressure on anything in their path.

That light quanta carries both momentum and energy can be shown using a vacuum
radiometer. In the vacuum radiometer it is the momentum of the quanta that causes the
vanes to spin. The energy of a photon is related to its frequency by the Planck equation.

It was not until the beginning of the 20 l century that the hypothesis that light quanta have
momentum was tested in a series of experiments by Nichols and Hull (1903) using a
silver mirror attached to a very sensitive torsion scale. In their experiment Nichols and
Hull took a transparent glass plate, silvered one side of it, hung the small silver mirror on
a torsion balance system and then shone a measured amount of light onto the mirror.

Deriving Einstein's Energy (4) n - Physics Initiative in Education
Expression E = mc 2

Nichols and Hull were able to confirm the relationship between the light pressure (P)
which was a measure of the light momentum absorbed per unit area by the mirror, the
rate of energy flow per unit area (S) and the speed of energy transport (c the speed of
light), namely

P = S/c (2)

The units of S is watts per square metre, of c is in metres per second and of P in Newtons
per square metre. The pressure P is just made up of contributions from many quanta. For
an individual light quanta, the momentum p of the light is related to its energy E by

momentum = p=E/c (3)

In their experiment they measured the energy flow per unit area (S) by absorbing the
incident light on a blackened disk and then measuring the disk's rate of temperature rise.

Sample calculation:

1. What is the light pressure P per square metre if S = 10,000 Watts /m and the

speed of light is 3.0 x 10 m/s?

P = 3.3x 10~ 5 N/m 2 .

2. A 10 tonne spacecraft is being moved through space by a solar sail. If the sail has

an area of 1 .0 x 1 6 m 2 and the
is the spacecraft accelerating?

an area of 1 .0 x 1 6 m 2 and the rate of energy flow per unit area is S = 10 W /m 2 , how fast

a = 3.3x 10" 6 m/s 2 .

Sidebar: Dr. Nichols died a rather sad and unexpected death. At an American Physical
Society meeting while on stage he stood to give a talk, spoke a few words and then sat
down and passed away of natural causes.

Deriving Einstein's Energy
Expression E = mc 2

(5) tz - Physics Initiative in Education

3^ A Simple Conceptual Derivation: Einstein's Box

The Cannon and the Projectile

Before we look at light quanta let us begin with a simple problem of mechanics, the
stopping of a projectile on a frictionless cart.

Consider a cannon of large mass mounted on a railway car. The railcar is free to move
without friction on the railway tracks. The cannon points towards the rear of the railcar
where a large similar mass is located. The second large mass at the rear of the railcar is
the catcher, designed to catch the projectile fired by the cannon (refer to Fig. 1 A Cannon
Mounted on a Railcar).

CKtZWEV-

=4

z

if

Fig. 1 A Cannon Mounted on a Railcar

Let the combined mass of the cannon and projectile equal mass M,

M

Mass

+ mass

projectile

(4)

Let, as well, the mass of the catcher equal the mass of the cannon and projectile (M).

Deriving Einstein's Energy (6) it - Physics Initiative in Education
Expression E = mc 2

It is convenient to assume that the combined mass of the cannon and catcher is
considerably larger than the mass of the railcar and projectile. The length of the railcar is
L. Evidently, the centre of mass of the system is at the centre of the railcar.

A small projectile of mass m, where m forms part of the combined mass (M) of the
cannon and projectile, is fired by the cannon towards the catcher. The projectile has
speed v. When the projectile is fired, by reaction (Newton's Third Law) the whole
railcar, including cannon and catcher (i.e. the system), is given a velocity V.

By the conservation of momentum

mv = 2 MV (5)

As soon as the projectile impacts the catcher it transfers its momentum to the system,
bringing the system to a stop. By what distance has the railcar shifted along the tracks?
The system has moved with velocity V given by

V = mv/2M (6)

The time of flight of the projectile is just the length of the railcar L divided by V.

A t = L/v (7)

The shift or movement of the system is the velocity times the time of flight, namely

d shift = V A t = V ( L/v) = Lm/2M (8)

Notice that the velocity of the projectile v drops out of this expression.

What has happened to the centre of mass of the system? Before the shot, because of the
symmetrical distribution of the masses, the centre of mass was evidently at the centre of

Deriving Einstein's Energy (7) jt - Physics Initiative in Education
Expression E = mc 2

the railcar. After the shot the railcar has shifted a distance; furthermore, the cannon has
lost and the catcher has gained mass projectile the mass of the projectile.

We can express the shift by the expression

(M - m) (L/2 + d shift ) = (M + m)(L/2 - d shlft ) (9)

A little algebra will show that even though the projectile has traveled the length of the
railcar the centre of mass of the system has remained fixed or stationary, with
respect to an outside observer. This means then that

Md-m(L/2) =0 (10)

Think about it: Why has the centre of mass of the system remained unchanged?

There is a general theorem in mechanics that states that the centre of mass of a system
that is at rest will remain at rest as long as no external forces are acting on the system.
Being an isolated system with no external forces acting on it, the centre of mass of the
railcar, cannon, catcher and projectile system must remain stationary. Note this is the
same thing as saying that the total momentum of the box remains zero.

The fact that the centre of mass of a system with no external forces acting remains
stationary will be used in the next section to derive Einstein's Mass-Energy Equation.
Presupposing that light consists of quanta and that each quanta is endowed with a
momentum p = E / c, (E is the energy of the quanta), we shall apply essentially the same
argument used with the cannon and the railcar to the Emission and Absorption of light.

Deriving Einstein's Energy
Expression E = mc 2

(8)

% - Physics Initiative in Education

Einstein's Box

Consider a box which emits a quanta of light energy from one end (A) of mass M which
is absorbed by the other end (B) of mass M. The box has a total mass 2M and length L.
The quanta of light energy carries momentum p = E / c.

Since the box is isolated and is not subjected to any external forces, the centre of mass of
the box and therefore the total momentum of the box must remain zero (refer to Fig. 2
Einstein's Box - Before and After the Quanta Emission).

A -

Center of Mis*

Fig. 2. Einstein's Box - Before and After the Quanta Emission

After emitting the light quanta (also known as a photon) the box recoils with velocity

v= E/2Mc

(11)

After traveling for a short time At the photon hits the other end of the box and conveys an
impulse equal and opposite to the initial impulse. Thus the box moves through a distance

d = v At = E L/ (2Mc 2 )

(12)

Deriving Einstein's Energy (9) it - Physics Initiative in Education
Expression E = mc 2

But this being an isolated system, the centre of mass of the box must remain unchanged,
or stationary with respect to an outside observer. After the emission and absorption of
the quanta of light energy the centre of mass is located by

(M - (x) (L/2 + d) = (M + n)(L/2 - d ) (13)

We therefore postulate that the recoil of the photon radiation carried with it the an
"equivalent mass" [i such that

Md -nL/2 = (14)

Replacing d in this equation with the expression for the distance through which the box
moved allows us to compute the "equivalent mass" \i to the light quanta energy E on the
basis of the theorem that the centre of mass of the system remains unchanged with
respect to the outside observer. This means then that the "equivalent mass" and the
energy of the light quanta must be related by

[i =E/c 2 (15)

Note that the length L and mass 2 M of the box drop out of the expression. We have then

E = ^c 2 (16)

There is something very important to note about the "equivalent mass" |i of the quanta.
Light quanta, and any particles that travel at the speed of light in vacuum, are massless
particles. That is to say, light itself does not have mass. We can't stop it and put a
quanta of light on a scale. We can, on the other hand, watch light curve near a massive
object in space, as was predicted by General Relativity and later observed.

The derivation above is quite universal, and applies to any process that transfers energy.
We can therefore say that for massive particles, particles that travel slower than the speed

Deriving Einstein's Energy (10) it - Physics Initiative in Education
Expression E = mc 2

of light, they too have a similar equation that relates inertia mass m to their internal
energy E, namely

E = mc 2 (17)

which is Einstein's Mass-Energy Equation. A further explanation of the Einstein Box
can be found in the book by Max Born titled "Einstein's Theory of Relativity."

4, Relativistic Mass Derivation

Underlying Einstein's Mass-Energy Equation is the concept that mass, like relativistic
time and length, is relative. Both time and length depends on the relative speed of the
object observed, and so does the mass.

The relevant equation that relates the mass m in motion to the rest mass m of an object
moving at a constant speed v is

m= m /V(l-v 2 /c 2 ) (18)

If we multiply both sides of this equation by c then

mc 2 = m c 2 /V(l-v 2 /c 2 ) (19)

We can now take advantage of a mathematical fact that at small x values the inverse
square root of (1-x) can be expanded out, namely

l/V(l-x) ~ l+x/2 (20)

When we apply this approximation to the mass equation we get

mc 2 ~ m c 2 (l+ 1 / 2 v 2 /c 2 ) = m c 2 + 1 / 2 m v 2 (21)

Deriving Einstein's Energy (11) jt - Physics Initiative in Education
Expression E = mc 2

The second term is the familiar kinetic energy term. The first term Einstein recognized as
new - in his own words "the mass of a body is a measure of its energy-content.'"

What this means is that if an object radiates (absorbs) off a certain amount of energy A E
its mass diminishes (increases) by an amount

A m = A E / c 2 (22)

In nuclear fission and fusion energy is released. Both the fission of massive nuclei such
as uranium-235 and the fusion of light nuclei like hydrogen have provided experimental
confirmation of Einstein's Mass - Energy Equation.

5, A Common Misconception

A force called the strong nuclear force holds the nucleus of atoms together. Past a certain
size, however, the electrostatic repulsion of the positive protons in the nucleus of U-235
and Pu-239 becomes greater than the combined strong nuclear force holding the nucleus
together. This is why the nuclei of atoms like Uranium-235 (U-235) or Plutonium-239
(Pu-239) are unstable. These large nuclei tend not to stay together and eventually fission.
A fission reaction involves the splitting of the nucleus of such an atom.

These nuclei will either fission on their own accord or can be induced to fission by
allowing a neutron to enter the nucleus. An induced fission occurs when a neutron enters
the nucleus of an atom, is captured by the nucleus, and causes the nuclei to break apart.

A fission reaction is many times more powerful than a chemical reaction. Where in a
chemical explosion around one electron volt of energy is released per chemical reaction,
about 200 million electron volts (200 Mev) is released per reaction in nuclear fission.
The fission of a single nuclei of Uranium-235 or Plutonium-239 is therefore over 200
million times more powerful per reaction than the most powerful chemical reaction.

Deriving Einstein's Energy (12) n - Physics Initiative in Education
Expression E = mc 2

A simple calculation will show that the fission of a kilogram of U-235 or Pu-239 will
release the same amount of energy as the combustion of 12,000 barrels of crude oil, 2,000
tonnes of thermal coal or 18 kilotonnes (kt) of the chemical explosive T.N.T..

A common misconception is that most of the energy released in nuclear fission comes
from the strong nuclear force and therefore Einstein's Mass - Energy Equation. In actual
fact the strong nuclear force only acts on the scale of a few femto metres (10" 15 m) and so
once the two halves of the fissioning nuclei are far enough apart the energy of coulombic
repulsion from the positive protons in the two halves is where the majority of the energy
released in a fission comes from.

The mass difference between the parent nucleus and the energy released from the
electrostatic repulsion from the positive protons in the two halves comes from the strong
nuclear force and therefore Einstein's Mass - Energy Equation.

Think About This: Estimate the energy released in the fission of a Uranium-235 atom.

An estimate of the size of a nucleus of atomic mass number A is R ~ 1.5 [A] 3 x 10" m
(A is raised to the third root). Imagine a U-235 nucleus which has 92 protons splits
exactly down the middle, with 46 protons in each half. Estimate the electromagnetic and
kinetic energy that is released as the two halves repel themselves apart.

You will find only a small a mass difference between the parent nucleus and the energy
released from the electrostatic repulsion from the positive protons in the two halves. This
comes from the strong nuclear force and therefore Einstein's Mass - Energy Equation.

If you are interested in learning more read Robert Serber's Book Los Alamos Primer, or
Richard Rhodes' book The Making of the Atom Bomb.

Deriving Einstein's Energy (13) jt - Physics Initiative in Education
Expression E = mc 2

6, The Sun: A Giant Fusion Reactor

We rely on the heat and the light from the sun to keep living things on earth alive and
well. The sun, which is a giant fusion reactor, radiates light energy at a prodigious rate,
about 3.7 xl0 26 W.

The average temperature of the surface of the earth is about 290 K. Were it not for the
heat and light from the sun, the surface of the earth would cool to just a few degrees
above absolute zero (the temperature of cold space).

It takes about 1 ,000 years for high-energy photons generated in the fusion reaction in the
core to diffuse their way to the surface of the sun. By the time these photons have arrived
at the surface of the sun they are more numerous in number and lower in energy.

While its surface temperature is about 5,800 K, deep at the core of the sun a temperature
in excess of several millions degrees is reached. At the core hydrogen is being fused in a
multi-step process making helium. Each time 4 hydrogen atoms are fused into a helium

i n

atom 4.28 x 10" J of energy is released.

Deriving Einstein's Energy (14) n - Physics Initiative in Education
Expression E = mc 2

Sample calculation:

1. If the sun radiates 3.7 x 10 26 J/s of energy and 4.28 x 10" 12 J of energy is released

every time 4 hydrogen atoms are fused into a helium atom, estimate the number of
reactions occurring per second at the core of the sun.

TO

number of reactions per second ~ 3.5 x 10 reactions/s

2. If the sun radiates 3.7 x 10 26 J of energy per second estimate the number of tonnes

of mass the sun is losing per second (use Einstein's Mass - Energy Equation)

mass loss = 3.7 x 10 26 J/s /(3.0 x 10 8 m/s) 2 = 4.1 x 10 6 Tonnes/ second

3. If the sun is about 6 billion years old and has been using up mass at the rate

outlined in question 2, how many earth masses of hydrogen have been used up over that
period? (the mass of the earth is about 6x10 kg)

number of earth masses -130

Deriving Einstein's Energy (15) it - Physics Initiative in Education
Expression E = mc 2

Pion Decay and Relativistic Energy

Pions are the particles that glue together the protons and the neutrons within a nucleus. If
you bombard a target nucleus with high energy protons you produce pions. At TRIUMF
hundreds of pions are being produced each second as protons collide with targets. These
pions escape from the nucleus and travel a short distance before they decay. A pion is a
form of meson, and meson physics is a central function of the Tri-University Meson
Facility (notice the term meson in TRIUMF 's name).

Consider the decay of a positive pion into a positive muon and neutrino. The muon also
has a mass, while the neutrino has neither charge nor a mass. Neutrino means little
neutral one. The kinematics diagram for the decay in the Laboratory Frame is

>-

£&

Let us analyze the decay of the positive pion using Grade 12 physics. We know both
Energy and Momentum must be conserved.

In the laboratory frame, which is where we observe, the Conservation of Energy requires

E K =Ev +E^= Ev + V(Pu. 2 + rmi 2 ) (23)

where mjj, is the rest mass of the muon, and P|i is the muon's momentum. E^ is the total
energy of the pion and Ev is the energy of the neutrino.

Deriving Einstein's Energy (16) it - Physics Initiative in Education
Expression E = mc 2

Recall for relativistic particles with mass that

E = V (P 2 + m 2 ) (24)

For a mass-less particle like the neutrinos Ev = Pv. (we use the convention of setting c,
the speed of light equal, to 1. The units for c can be added back at the end of the
calculation).

The Conservation of Momentum requires for the horizontal and vertical directions that

P* = V (E n 2 - m K 2 ) = Pv cos 9v + P|i cos 9|i (25)

0= Pv sinBv + P|isin9|a (26)

Rewriting the first equation and squaring both sides gives

(P n - Ev cos 9v f = Py 2 cos 2 0fx (27)

We have used the fact that for a mass-less particle like the neutrinos Ev = Pv . Similarly
for the second equation

Ev 2 sin 2 9v = P^i 2 sin 2 9^ (28)

We can use these equations to solve for the energy of the neutrino, Ev, in the Laboratory
Frame. Adding these two equations together give

P„ 2 - 2 P„ Ev cos 9v + Ev 2 cos 2 9v + Ev 2 sin 2 9v = P|/ cos 2 9fx + P^i 2 sin 2 9^ (29)

which we can simplify to

Deriving Einstein's Energy (17) n - Physics Initiative in Education
Expression E = mc 2

E n 2 -m n 2 -2P„ Ev cosBv + Ev 2 = P^ 2 (30)

If we now take the Conservation of Energy equation, rewrite it and square both sides we
get

(E* - Ev) 2 =(Pn 2 + mu 2 ) (31)

Expanding the left side of this equation out ,and moving ni|a 2 to the left side, we find

E K 2 - 2 E K Ev + Ev 2 - mu 2 = P^i 2 (32)

Now we have two equations that equal the same thing, namely P|i . So the two equations
should be equal to each other, namely

E I 2 -m„ 2 -2P 1I Ev cosBv + Ev 2 = E^ 2 -2E„Ev + Ev 2 - mu 2 (33)

Gathering the mass terms on the right hand side and simplifying the equation gives

2Ev (E»- P„ cosOv) = m n 2 - m\i 2 (34)

Notice that the right hand side is a difference of the square of two rest masses. Rest
masses are fixed measures. They do not change from frame of reference to frame of
reference. We will use this to our advantage in a moment.

In the Laboratory Frame the energy of the neutrino Ev is given by

Ev = (m, 2 - mu 2 ) / 2 (E, - P* cos 9v ) (35)

We could have also done the calculation in the Centre of Mass frame, where the pion is
initially at rest and the neutrino and muon fly off in opposite directions.

Deriving Einstein's Energy
Expression E = mc 2

(18) it - Physics Initiative in Education

A kinematics diagram for the pion decay in the Centre of Mass frame is

t

Ev*

hecifH^fl

<i ta^ pi<W

^ */* V*UO*i

In this case P,t* = and E^* = m n , where the star labels them as measurements made in
the Centre of Mass Frame. So then

2 m K Ev* = m„ - m|i

(36)

or

Ev* = ( m, 2 - m|i 2 ) / m,

(37)

This is an interesting expression in that the right hand side is made up of purely invariant
measures.

Transformation of the Energy Expression from one Frame to the Next.

Using the fact that we now have two expressions, one from the Lab Frame and one from
the Centre of Mass frame, which equal to a difference of the square of two rest masses,
we can find a relationship between the two frames.

Deriving Einstein's Energy (19) it - Physics Initiative in Education
Expression E = mc 2

In the Laboratory Frame

2 Ev (E„ - P,! cos 0v ) = m, 2 - mjj 2 (38)

and in the Centre of Mass frame

2 m K Ev* = m, 2 - m\x 2 (39)

so then

Ev* = Ev (E, - P„ cos 9v ) / m, (40)

If we remember that from Grade 1 1 that E„ I m n = y, and ¥„ / m, = p y, then for the mass-
less neutrino

Ev* = y(l-(3cos9v)Ev (41)

where |3 = v/c and y = 1 / V( 1 - (3 2 ), and v is the velocity the Laboratory Frame with
respect to the Centre of Mass Frame. This expression, with Ev* on one side and Ev on the
other side, is the Transformation equation that relates the energy of the neutrino Ev* in
the Centre of Mass frame to the energy of the neutrino Ev in the Laboratory Frame.

It is easily shown that if the particle has mass then the energy transformation expression
between two frames in relative motion, is given by

E * = y (E - P cos 9v ) (42)

Deriving Einstein's Energy (20) n - Physics Initiative in Education
Expression E = mc 2

The n - Physics Initiative in Education is a program begun in support of the UN
sponsored 2005 International Year of Physics, and Physics Education in the Province of
British Columbia.

This n - Physics Initiative in Education Module may be copied and used freely with
attribution to the source and the University of British Columbia. © 2005

Additional modules covering topics of interest to middle and high school students in
British Columbia will be made available through the Department of Physics and
Astronomy at the University of British Columbia during the period 2005 -2010.

Should you require assistance in the development or presentation of physics curriculum at
a middle and high school please feel free to contact the Department of Physics and
Astronomy at the University of British Columbia.

This module was prepared by:

Patrick Bruskiewich,
M.Sc, B.Sc, B.Ed., BCCT

Department of Physics and Astronomy - UBC
email: patrickb@physics.ubc.ca

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